To find the flux through the shaded face of the cube, we use symmetry:

1. The total flux through a closed surface is given by Gauss's law:
\[
\Phi_{\text{total}} = \frac{q}{\varepsilon_0}.
\]

2. The charge \(q\) is at one vertex of the cube. It is shared equally among 8 cubes because the vertex belongs to 8 adjacent cubes. Hence, the charge effectively enclosed in one cube is:
\[
q_{\text{enclosed}} = \frac{q}{8}.
\]

3. The flux through the entire surface of one cube is:
\[
\Phi_{\text{cube}} = \frac{q_{\text{enclosed}}}{\varepsilon_0} = \frac{q}{8\varepsilon_0}.
\]

4. A cube has 6 faces, and due to symmetry, the flux through each face is the same. Therefore, the flux through one face is:
\[
\Phi_{\text{one face}} = \frac{\Phi_{\text{cube}}}{6} = \frac{q}{48\varepsilon_0}.
\]

5. The shaded face is shared by 2 adjacent cubes, so it will receive twice the flux compared to one face:
\[
\Phi_{\text{shaded face}} = 2 \times \frac{q}{48\varepsilon_0} = \frac{q}{24\varepsilon_0}.
\]

Thus, the flux through the shaded face is:

\[
\Phi = \frac{q}{24\varepsilon_0}.
\]

To solve the problem using the formula:

\[
\Phi = \frac{Q}{2\varepsilon_0} \left(1 - \cos\theta\right),
\]

where \(\theta\) is the half-angle subtended by the disc at the charge, follow these steps:

1. The flux through the disc is given to be one-fourth of the total flux:
\[
\Phi = \frac{1}{4} \cdot \frac{Q}{\varepsilon_0}.
\]

Substituting this into the formula:
\[
\frac{1}{4} \cdot \frac{Q}{\varepsilon_0} = \frac{Q}{2\varepsilon_0} \left(1 - \cos\theta\right).
\]

2. Simplify:
\[
\frac{1}{4} = \frac{1}{2} \left(1 - \cos\theta\right).
\]

3. Multiply through by 2:
\[
\frac{1}{2} = 1 - \cos\theta.
\]

4. Solve for \(\cos\theta\):
\[
\cos\theta = 1 - \frac{1}{2} = \frac{1}{2}.
\]

5. Using the geometry, \(\cos\theta = \frac{a}{\sqrt{a^2 + R^2}}\). Substituting:
\[
\frac{a}{\sqrt{a^2 + R^2}} = \frac{1}{2}.
\]

6. Square both sides:
\[
\frac{a^2}{a^2 + R^2} = \frac{1}{4}.
\]

7. Rearrange:
\[
4a^2 = a^2 + R^2.
\]

\[
3a^2 = R^2.
\]

8. Solve for \(a\):
\[
a = \frac{R}{\sqrt{3}}.
\]

Thus, the relation is:

\[
a = \frac{R}{\sqrt{3}}.
\]

\phi= \frac{q}{6\varepsilon_{0}} + \frac{q}{24\varepsilon_{0}}= \frac{5q}{24\varepsilon_{0}}

As number of electric field lines passing through both the surfaces is equal electric flux associated through both of them is equal.

The area vector \( \vec{A} \) for a surface in the \( YZ \)-plane points along the \( \hat{i} \)-direction, with magnitude \( A = 20 \). Thus,
\[
\vec{A} = 20\hat{i}.
\]

The electric field is given by:
\[
\vec{E} = 5\hat{i} + 4\hat{j} + 9\hat{k}.
\]

Flux \( \Phi \) is:
\[
\Phi = \vec{E} \cdot \vec{A} = (5\hat{i} + 4\hat{j} + 9\hat{k}) \cdot 20\hat{i}.
\]

Only the \( \hat{i} \)-component contributes:
\[
\Phi = 5 \times 20 = 100 \, \text{units}.
\]

Specifying the total charge inside the surface is a sufficient condition for finding the electric flux \( \Phi \) through a closed surface.

According to Gauss's law:
\[
\Phi = \frac{q_{\text{enclosed}}}{\varepsilon_0}
\]

The flux depends only on the enclosed charge \( q_{\text{enclosed}} \), regardless of the surface's shape or the charge distribution.

Since the electric field \( E \) is parallel to the cylinder's axis, the field lines enter and exit symmetrically through the two flat circular ends. The curved surface has no perpendicular component of the field, so no flux passes through it.

By Gauss's law, the net flux through the entire surface is:
\[
\Phi = \text{Charge enclosed} / \varepsilon_0
\]

As no charge is enclosed, \( \Phi = 0 \).

The electric flux through the hemisphere is due to the uniform electric field passing perpendicularly through the circular plane of radius \( R \).

Flux through the circular plane:
\[
\Phi = E \cdot \text{Area of circular plane} = E \cdot \pi R^2
\]

Since no flux passes through the curved surface (field is parallel to it), the total flux is:
\[
\Phi = \pi R^2 E
\]

Gauss's law states that the total electric flux passing through a closed surface is equal to the charge enclosed within the surface divided by the permittivity of the medium.

In short: The net electric field around a closed surface depends on the charge inside it.

By Gauss's law, the total flux through the cube is \(\Phi = \frac{q_{\text{enclosed}}}{ε_0}\). Since the charge is at the center, the flux through one of the six faces is \(\Phi_1 = \frac{\Phi}{6} = \frac{Q \times 10^{-6}}{6ε_0}\).